Question

The internal diameter of a hollow iron sphere is 14cm and external diameter is 28cm if the density of iron is 7800kg/m3 what is the the mass of the sphere?

Answer 1

Given :

◾Inner diameter =14 cm

◾External diameter = 28 cm

◾Density of iron = 7800kgm¯³

To Find :

◾Mass of the sphere

Formula used :

$Volume \: of \: sphere = \frac{4}{3} \pi {r}^{3}$

Solution :

Internal radius (r) = 7 cm

External radius (R) = 14 cm

Volume of hollow iron sphere = [Internal volume of sphere - External volume of sphere]

$Volume \: of \: hollow \: iron \: sphere = ( \frac{4}{3}\pi {R}^{3} - \frac{4}{ 3} \pi {r}^{3} ) \\ Volume \: of \: hollow \: iron \: sphere = \frac{4}{3}\pi(R^{3} - {r}^{3} ) \\ Volume \: of \: hollow \: iron \: sphere = \frac{4}{3} \times \frac{22}{7} ( {14}^{3} - {7}^{3} ) \\ Volume \: of \: hollow \: iron \: sphere = \frac{88}{21} \times 2401 \\ Volume \: of \: hollow \: iron \: sphere = 10061.34 {cm}^{3}$

As we know, Density = mass /Volume

So, mass = Density × volume

Conversion :

1 m³ = 1000000 cm³

1 cm ³ = 1/1000000 m³

10061.34 cm ³ = 10061.34/1000000 m³

10061.34 cm ³ = 0.01006134 m³

Therefore,

Mass of sphere = 7800×0.01006134 kg

Mass of sphere = 78.47 kg